Multiple choice

If one root of the equation $z^2+(a+i)z+b+ic=0$ is real, when a, b$\epsilon$ R, then $c^2+b-ac$ is equal to.

  1. $0$
  2. $-1$
  3. $1$

  4. None of these
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A Correct answer
Explanation

Let the real root be r. Then r^2 + (a+i)r + b + ic = 0 => (r^2 + ar + b) + i(r + c) = 0. For this to be zero, both real and imaginary parts must be zero. r + c = 0 => r = -c. Substitute into the real part: (-c)^2 + a(-c) + b = 0 => c^2 - ac + b = 0.

AI explanation

Let the real root of the equation z^2 + (a+i)z + b+ic = 0 be k. Since k is a real number, the imaginary part of the equation must be zero. This means the imaginary component k + c = 0, so k = -c. The real part gives the equation k^2 + ak + b = 0. Substituting k = -c into the real part gives c^2 - ac + b = 0. Rearranging this equation gives c^2 + b - ac = 0.

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